大家希望能选出一个人,所有人都认识他,但同时他不认识镇上除自己以外的其他人(在此,我们默认每个人自己认识自己)。可是小镇里的人太多了,一下子大家谁也说服不了谁
第一种方法
include "stdafx.h"
#include<iostream>
#include<vector>
#include<string>
using namespace std;
int main()
{
int num;
cin >> num;
//vector<vector<int>> heads;
while (num!=0)
{
num--;
int n;
int m;
cin >> n >> m;
vector<int> inVec(n+1,1);
vector<int> outVec(n + 1, 1);
vector<int> head;
for (int i = 0;i < m;i++)
{
int a, b;
cin >> a >> b;
if (a == b)
{
continue;
}
else
{
outVec[a]++;
inVec[b]++;
}
}
for (int i = 1;i < n + 1;i++)
{
if (inVec[i] == n && outVec[i] == 1)
{
head.push_back(i);
}
}
if (head.size() == 0)
{
cout << 0 << endl << endl;
}
else
{
cout << head.size() << endl;
for (int i = 0;i < head.size() - 1;i++)
{
cout << head[i] << " ";
}
cout << head[head.size() - 1] << endl;
}
}
return 0;
}
第二种:
#include "stdafx.h"
#include<iostream>
#include<vector>
#include<string>
using namespace std;
//所有的人都认识他
bool getAllpeople(vector<vector<int>> relation, int people)
{
bool result = true;
for (int x = 1;x < relation.size();x++)
{
if (relation[x][people] != 1)
{
result = false;
break;
}
}
return result;
}
//他不认识所有的人
bool getNopeope(vector<vector<int>> relation, int people)
{
bool result = true;
for (int y = 1;y < relation.size();y++)
{
if (y == people)
{
continue;
}
if (relation[people][y] == 1)
{
result = false;
break;
}
}
return result;
}
int main()
{
int num;
cin >> num;
while (num != 0)
{
num--;
int n;
int m;
cin >> n >> m;
vector<vector<int>> relation(n + 1, vector<int>(n + 1, 0));
vector<int > head;
for (int i = 0;i < m;i++)
{
int a, b;
cin >> a >> b;
relation[a][b] = 1;
}
for (int i = 1;i < relation.size();i++)
{
relation[i][i] = 1;
}
for (int people = 1;people <relation.size();people++)
{
if (getAllpeople(relation, people) == true && getAllpeople(relation, people) == true)
{
head.push_back(people);
}
}
if (head.size() == 0)
{
cout << 0 << endl << endl;
}
else
{
cout << head.size() << endl;
for (int j = 0;j < head.size() - 1;j++)
{
cout << head[j] << " ";
}
cout << head[head.size() - 1] << endl;
}
}
return 0;
}
评论:第一种方法能够通过,第二种不能通过,因为第二种的占用的空间多